Question

A buffer solution contains 0.72 mol of hydrosulfuric acid (H2S) and 0.70 mol of sodium hydrogen sulfide (NaHS) in 8.60 L. The Ka of hydrosulfuric acid (H2S) is Ka = 9.5e-08.

(a) What is the pH of this buffer?

pH =

(b) What is the pH of the buffer after the addition of 0.17 mol of NaOH? (assume no volume change)

pH =

(c) What is the pH of the original buffer after the addition of 0.53 mol of HI? (assume no volume change)

pH =

Answer 1

Given that a buffer solution contains 0.72 mol of hydrosulfuric acid (H2S) and 0.70 mol of sodium hydrogen sulfide (NaHS) in 8.60 L.

The Ka of hydrosulfuric acid (H2S) is Ka = 9.5 x 10^-8

[H2S] = moles/ volume = 0.72 mol/ 8.60 L= 0.72/8.60 M

[NaHS] = moles/ volume = 0.70 mol/ 8.60 L= 0.70/8.60 M

a)

pH = -logKa + log [NaHS]/ [H2S]

= -log (9.5 x 10^-8) + log ( 0.72/8.60 / 0.70/8.60)

= 7.03

pH = 7.03

Therefore, pH of the buffer = 7.03

b) pH of the buffer after the addition of 0.17 mol of NaOH

H2S + NaOH ---------------> NaHS + H2O

0.72 mol 0.17 mol 0

----------------------------------------------------------------------------

0.72-0.17 0 0.17 mol

= 0.55 mol

Hence,

[H2S] = 0.55 mol

[NaHS] = initial moles + 0.17 mol = 0.70 mol + 0.17 mol = 0.87 mol

pH = -logKa + log [NaHS]/ [H2S]

= -log (9.5 x 10^-8) + log ( 0.87 / 0.55)

= 7.22

pH = 7.22

Therefore, pH of the buffer = 7.22

c) pH of the original buffer after the addition of 0.53 mol of HI

NaHS + HI ---------------> H2S + NaI

0.70 mol 0.53 mol 0

----------------------------------------------------------------------------

0.70-0.53 0 0.53 mol

= 0.17 mol

Hence,

[NaHS] = 0.17 mol

[H2S] = initial moles + 0.53 mol = 0.72 mol + 0.53 mol = 1.25 mol

pH = -logKa + log [NaHS]/ [H2S]

= -log (9.5 x 10^-8) + log ( 0.17 / 1.25)

= 6.15

pH = 6.15

Therefore, pH of the buffer = 6.15